Problem: There is a light source on the plane. This source is so small that it can be represented as point. The light source is moving from point (a, s_y) to the (b, s_y) (s_y < 0) with speed equal to 1 unit per second. The trajectory of this light source is a straight segment connecting these two points. 

There is also a fence on OX axis represented as n segments (l_i, r_i) (so the actual coordinates of endpoints of each segment are (l_i, 0) and (r_i, 0)). The point (x, y) is in the shade if segment connecting (x,y) and the current position of the light source intersects or touches with any segment of the fence.

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You are given q points. For each point calculate total time of this point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y).

Input

First line contains three space separated integers s_y, a and b (-10^9 ≤ s_y < 0, 1 ≤ a < b ≤ 10^9) — corresponding coordinates of the light source.

Second line contains single integer n (1 ≤ n ≤ 2 ⋅ 10^5) — number of segments in the fence.

Next n lines contain two integers per line: l_i and r_i (1 ≤ l_i < r_i ≤ 10^9, r_{i - 1} < l_i) — segments in the fence in increasing order. Segments don't intersect or touch each other.

Next line contains single integer q (1 ≤ q ≤ 2 ⋅ 10^5) — number of points to check.

Next q lines contain two integers per line: x_i and y_i (1 ≤ x_i, y_i ≤ 10^9) — points to process.

Output

Print q lines. The i-th line should contain one real number — total time of the i-th point being in the shade, while the light source is moving from (a, s_y) to the (b, s_y). The answer is considered as correct if its absolute of relative error doesn't exceed 10^{-6}.

Example

Input

-3 1 6
2
2 4
6 7
5
3 1
1 3
6 1
6 4
7 6


Output

5.000000000000000
3.000000000000000
0.000000000000000
1.500000000000000
2.000000000000000

Note

  * The 1-st point is always in the shade; 
  * the 2-nd point is in the shade while light source is moving from (3, -3) to (6, -3); 
  * the 3-rd point is in the shade while light source is at point (6, -3). 
  * the 4-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and at point (6, -3); 
  * the 5-th point is in the shade while light source is moving from (1, -3) to (2.5, -3) and from (5.5, -3) to (6, -3); 

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
using namespace std;
inline long double max1(long double a, long double b) {
  if (a < b) return b;
  return a;
}
inline long double min1(long double a, long double b) {
  if (a < b) return a;
  return b;
}
int main() {
  long double s, a, b;
  scanf("%Lf %Lf %Lf", &s, &a, &b);
  s = -s;
  long double n, sum = 0;
  scanf("%Lf", &n);
  vector<long double> low(n), high(n);
  vector<long double> presum(n, (long double)0);
  for (int i = 0; i < n; ++i) {
    scanf("%Lf %Lf", &low[i], &high[i]);
    presum[i] = (high[i] - low[i]);
    if (i > 0) presum[i] = presum[i] + presum[i - 1];
  }
  int q, tt = 0;
  scanf("%d", &q);
  while (q--) {
    tt++;
    long double x, y;
    scanf("%Lf %Lf", &x, &y);
    long double lower = ((long double)((x - a) * y)) / (s + y);
    if (lower < 0) lower = -lower;
    if (a < x)
      lower = x - lower;
    else
      lower = x + lower;
    long double higher = ((long double)((x - b) * y)) / (s + y);
    if (higher < 0) higher = -higher;
    if (b < x)
      higher = x - higher;
    else
      higher = x + higher;
    int leftind = lower_bound((low).begin(), (low).end(), lower) - low.begin();
    if (leftind - 1 > -1 && high[leftind - 1] > lower) leftind--;
    int rightind =
        lower_bound((low).begin(), (low).end(), higher) - low.begin();
    rightind--;
    long double tempsum = 0;
    if (leftind == low.size() || rightind == -1 || rightind < leftind)
      tempsum = 0;
    else if (leftind == rightind)
      tempsum = max1(min1(high[leftind], higher) - max1(low[leftind], lower),
                     (long double)0);
    else {
      tempsum = tempsum + max1(high[leftind] - max1(lower, low[leftind]) +
                                   min1(higher, high[rightind]) - low[rightind],
                               (long double)0);
      tempsum = tempsum + presum[rightind - 1] - presum[leftind];
    }
    long double d =
        max1(((long double)((s + y) * tempsum)) / y, (long double)0);
    printf("%.15Lf\n", d);
  }
}